### Author Topic: Complex eigenvalues  (Read 1550 times)

#### Karim El Komaishy

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##### Complex eigenvalues
« on: November 07, 2018, 11:19:35 PM »
When I tried some of the questions that involved matrices with complex eigenvalues, I sometimes ended up with matrices whose reduced row echelon forms that have pivot positions on each column (after subracting the main diagonal entries with the eigenvalues; no free variables). Does this sometimes happen whenever I try to find eigenvectors with this?

#### Victor Ivrii

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##### Re: Complex eigenvalues
« Reply #1 on: November 08, 2018, 03:31:50 AM »
Not c lear what do you mean.May be an example?

#### Jingyi Wang

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##### Re: Complex eigenvalues
« Reply #2 on: November 13, 2018, 11:39:39 AM »
Not really I think. We have to get a matrix contains non-zero number on the first row and zero on the second row in reduced row echelon forms.

#### Tzu-Ching Yen

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##### Re: Complex eigenvalues
« Reply #3 on: November 13, 2018, 12:00:23 PM »
There should always be at least one free variable if eigenvalue is chosen correctly. Free variable need to be present so that your solution is a line or plane or etc.

#### Victor Ivrii

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##### Re: Complex eigenvalues
« Reply #4 on: November 13, 2018, 12:42:46 PM »
If the characteristic polynomial $\det (A-kI)$ has a multiple root (say root $k$ of multiplicity $m$) then it can have any number $l=1,\ldots, m$ of linearly independent eigenvectors, associated with this eigenvalue (we do not distinguish here real or complex eigenvalues). Then one needs to find Jordan (echelon?) form.

In particular, let $m=2$. Then either there are two l.i. eigenvectors $\mathbf{e}_1$, $\mathbf{e}_2$; then the corresponding part of the solution is
$$e^{kt}\Bigl( C_1 \mathbf{e}_1+C_2\mathbf{e}_2\Bigr)$$
or there exists a vector $\mathbf{f}$ s.t. $(A-k)^2\mathbf{f}=0$, but $\mathbf{e}=(A-k)\mathbf{f}\ne 0$; then the corresponding part of the solution is
$$e^{kt}\Bigl( C_1 \bigl[t\mathbf{e}+\mathbf{f}\bigr]+C_2\mathbf{e}\Bigr)$$
If $m=3$ then there are three cases: either there are three l.i. $\mathbf{e}_1$, $\mathbf{e}_2$, $\mathbf{e}_3$, in which case solution is similar to the former case of $m=2$, or there are a vector $\mathbf{f}$ s.t. $(A-k)^2\mathbf{f}=0$, but $\mathbf{e}_1=(A-k)\mathbf{f}\ne 0$, and $\mathbf{e}_2$, also eigenvector but not proportional $\mathbf{e}_1$; then the corresponding part of the solution is
$$e^{kt}\Bigl( C_1 \bigl[t\mathbf{e}_1+\mathbf{f}\bigr]+C_2\mathbf{e}_1+C_3\mathbf{e}_2\Bigr);$$
or there exists a vector $\mathbf{g}$ s.t. $(A-k)^3\mathbf{g}=0$, but $\mathbf{e} =(A-k)^2\mathbf{g}\ne 0$; let $\mathbf{f} =(A-k)\mathbf{g}$ and
the corresponding part of the solution is
$$e^{kt}\Bigl( C_1 \bigl[\frac{t^2}{2}\mathbf{e}_1+t\mathbf{f}+\mathbf{h}\bigr]+C_2\bigl[t\mathbf{e} +\mathbf{f}\bigr]+C_3\mathbf{e}\Bigr).$$
and so on...